Question

The value of $k \in \mathbb{N}$ for which the integral \[ I_n = \int_0^1 (1 - x^k)^n \, dx, \, n \in \mathbb{N}, \] satisfies $147 \, I_{20} = 148 \, I_{21}$ is:

• A. 10
• B. 8
• C. 14
• D. 7

Answer 1

The Correct Option is D

\[ I_n = \int_{0}^{1} (1 - x^k)^n \cdot 1\, dx \] \[ I_n = (1 - x^k)^n \cdot x - nk \int_{0}^{1} (1 - x^k)^{n-1} \cdot x^{k-1}\, dx \] \[ I_n = nk \left[ \int_{0}^{1} \left( (1 - x^k)^n - (1 - x^k)^{n-1} \right) dx \right] \] \[ I_n = nk I_n - nk I_{n-1} \] \[ \frac{I_n}{I_{n-1}} = \frac{nk}{nk + 1} \] \[ \frac{I_{21}}{I_{20}} = \frac{21k}{1 + 21k} \] \[ = \frac{147}{148} \Rightarrow k = 7 \] 

Answer 2

The given integral is:
\[ I_n = \int_0^1 (1 - x^k)^n dx. \]
Using integration by parts, we get:
\[ I_n = \frac{nk}{nk + 1} I_{n-1}. \]
Iterating this formula, the relationship becomes:
\[ \frac{I_n}{I_{n-1}} = \frac{nk}{nk + 1}. \]
Given:
\[ \frac{I_{21}}{I_{20}} = \frac{147}{148}, \]
we substitute into the formula:
\[ \frac{21k}{21k + 1} = \frac{147}{148}. \]
Cross-multiplying and solving:
\[ 148 \cdot 21k = 147 \cdot (21k + 1), \]
\[ 148 \cdot 21k = 147 \cdot 21k + 147, \]
\[ 21k = 147 \implies k = 7. \]